Question

12£®ÔÚƽÃæÖ±½Ç×ø±êϵxOyÖУ¬Ö±ÏßlµÄ²ÎÊý·½³ÌΪ$\left\{\begin{array}{l}{x=3-\frac{\sqrt{2}}{2}t}\\{y=\sqrt{5}+\frac{\sqrt{2}}{2}t}\end{array}\right.$£¨tΪ²ÎÊý£©£®ÔÚÒÔÔ­µãOΪ¼«µã£¬xÖáÕý°ëÖáΪ¼«ÖáµÄ¼«×ø±êÖУ¬Ô²CµÄ·½³ÌΪ¦Ñ=2$\sqrt{5}$sin¦È£®
£¨1£©Ð´³öÖ±ÏßlµÄÆÕͨ·½³ÌºÍÔ²CµÄÖ±½Ç×ø±ê·½³Ì£»
£¨2£©ÉèµãP£¨3£¬$\sqrt{5}$£©£¬Ö±ÏßlÓëÔ²CÏཻÓÚA¡¢BÁ½µã£¬Çó$\frac{1}{|PA|}$+$\frac{1}{|PB|}$µÄÖµ£®

Answer 1

·ÖÎö £¨1£©°ÑÖ±ÏßlµÄ²ÎÊý·½³ÌÏûÈ¥²ÎÊýt¿ÉµÃ£¬ËüµÄÖ±½Ç×ø±ê·½³Ì£»°ÑÔ²CµÄ¼«×ø±ê·½³ÌÒÀ¾Ý»¥»¯¹«Ê½×ª»¯ÎªÖ±½Ç×ø±ê·½³Ì£®
£¨2£©°ÑÖ±ÏßlµÄ²ÎÊý·½³Ì´úÈëÔ²CµÄÖ±½Ç×ø±ê·½³Ì£¬µÃ  ${£¨{3-\frac{{\sqrt{2}}}{2}t}£©^2}+{£¨{\frac{{\sqrt{2}}}{2}t}£©^2}=5$£¬¼´${t^2}-3\sqrt{2}t+4=0$£¬½áºÏ¸ùÓëϵÊýµÄ¹Øϵ½øÐнâ´ð£®

½â´ð ½â£º£¨1£©ÓÉ$\left\{\begin{array}{l}x=3-\frac{{\sqrt{2}}}{2}t\\ y=\sqrt{5}+\frac{{\sqrt{2}}}{2}t\end{array}\right.$µÃÖ±ÏßlµÄÆÕͨ·½³ÌΪ$x+y-3-\sqrt{5}=0$£®
ÓÖÓÉ$¦Ñ=2\sqrt{5}sin¦È$µÃÔ²CµÄÖ±½Ç×ø±ê·½³ÌΪ${x^2}+{y^2}-2\sqrt{5}y=0$
¼´${x^2}+{£¨y-\sqrt{5}£©^2}=5$£»
£¨2£©°ÑÖ±ÏßlµÄ²ÎÊý·½³Ì´úÈëÔ²CµÄÖ±½Ç×ø±ê·½³Ì£¬
µÃ  ${£¨{3-\frac{{\sqrt{2}}}{2}t}£©^2}+{£¨{\frac{{\sqrt{2}}}{2}t}£©^2}=5$£¬¼´${t^2}-3\sqrt{2}t+4=0$£¬
ÓÉÓÚ$¡÷={£¨{3\sqrt{2}}£©^2}-4¡Á4=2£¾0$£¬
¹Ê¿ÉÉèt₁£¬t₂ÊÇÉÏÊö·½³ÌµÄÁ½ÊµÊý¸ù£¬
ËùÒÔ$\left\{{\begin{array}{l}{{t_1}+{t_2}=3\sqrt{2}}\\{{t_1}•{t_2}=4}\end{array}}\right.$£¬
¡àt₁£¾0£¬t₂£¾0¡­£¨7·Ö£©
ÓÖÓÐÖ±Ïßl¹ýµã$P£¨{3£¬\sqrt{5}}£©$£¬A¡¢BÁ½µã¶ÔÓ¦µÄ²ÎÊý·Ö±ðΪt₁£¬t₂
ËùÒÔ|PA|=t₁£¬|PB|=t₂
ËùÒÔ$\frac{1}{{|{PA}|}}+\frac{1}{{|{PB}|}}=\frac{1}{t_1}+\frac{1}{t_2}=\frac{{{t_1}+{t_2}}}{{{t_1}{t_2}}}=\frac{{3\sqrt{2}}}{4}$£®

µãÆÀ ±¾ÌâÖص㿼²éÁËÖ±ÏߵIJÎÊý·½³ÌºÍÆÕͨ·½³ÌµÄ»¥»¯¡¢¼«×ø±ê·½³ÌºÍÖ±½Ç×ø±ê·½³ÌµÄ»¥»¯¡¢Ö±ÏßÓëÔ²µÄλÖùØϵµÈ֪ʶ£¬ÊôÓÚÖеµÌ⣮